Two events in reference frame S occur 10 μs apart at the same ...

Question

Two events in reference frame S occur 10 μs apart at the same point in space. The distance between the two events is 2400 m in reference frame S'. What is the velocity of S' relative to S?

Accepted Answer

Hey, everyone in this problem, we're told that in reference frame s known as the earth station frame, the rocket launch in the signal reception occur at the same point in space but 15 microseconds apart in time in reference frame S prime known as the spacecraft frame, these events are 2800 m apart. We're asked to determine the velocity of S prime with respect to S we're given four answer choices. Option A 1.35 multiplied by C option B 0.528 multiplied by C option C 3.32 multiplied by C and option D 2.52 multiplied by C. So let's start by writing out the information we've been given. So we have frame S and we're told that in frame sk this is the earth frame, two events occur at the same point. If they occur at the same point, then delta X, the distance between them is just going to be zero and they occur 15 microseconds apart in time. So delta T is going to be equal to 15 microseconds. Now, for frame S prime, we're gonna look at those same variables are the corresponding variables. OK. Now, in S prime, we we're told the events are 2800 m apart. So delta X prime is going to be 2800 m and we aren't told any information about the time separation in delta T prime. Now, what we wanna find is the velocity of frame S prime with respect to S so recall that the velocity V of S prime with respect to X is going to be delta X prime divided by delta T prime. OK. Velocity is distance over time or displacement divided by time. Now we have delta X prime but we need delta T prime in order to calculate this value. OK. So we need this delta T prime value. Now let's call this equation. We've written V is equal to delta X prime divided by delta T prime equation star. And we're gonna come back to this once we figured out what delta T prime is. So now the question is how can we find delta T prime? Well, recall that space time intervals between two events is invariant. OK. So what that means is we have the following equation C squared multiplied by delta T squared minus delta X squared is equal to C squared multiplied by delta T prime squared minus delta X prime squared. OK. So the space time is the same in all of the inertial reference frame. OK. So we set them equal to each other. The space prime space time. Sorry. From frame S on the left, the spacetime from frame S prime. On the right now, on the left hand side, we have all these values, we have delta T and delta X, we now see that constant speed of light. On the right hand side, we have delta X prime. So the only unknown is delta T prime, we can substitute in our values and solve for delta T prime like we need. So doing that three multiplied by 10 to the exponent 8 m per second, all squared, OK. At speed of light squared multiplied by delta T squared. We're gonna convert this to our standard unit of seconds. So we have 15 multiplied by 10 to the exponent negative six seconds. That's all squared minus delta X squared. Well, that's just zero squared. So that term is gonna go away. We don't have to worry about it. This is going to be equal to three multiplied by 10 to the exponent eight meters per second. All squared multiplied by delta T prime squared minus 2800 m squared. All right. So let's go ahead and simplify again. We're solving for delta T simplifying on the left hand side, 2.025 multiplied by 10 to the exponent seven meters squared. Here we have meters per second squared multiplied by second squared. So those second squared will divide up and we're left with just meters squared. This is going to be equal to nine multiplied by 10 to the exponent 16 m per second multiplied by delta T prime squared. And sorry, this should be meters squared per second squared because we squared that speed of light and then we're gonna subtract 7.84 multiplied by 10 to the exponent 6 m squared. OK. Delta X prime squared. All right. So to isolate for T prime, we are first going to move this 7.84 multiplied by 10 to the exponent six to the left hand side by adding it. When we do that, we're left with nine, multiplied by 10 to the exponent 16 meter squared per second squared multiplied by delta T prime squared is equal to 2.809 multiplied by 10 to the exponent 7 m squared and dividing both sides by nine multiplied by 10 to the exponent 16. We get that delta T prime squared is equal to about 3.1211 repeated multiplied by 10 to the exponent negative 10 second squared. And the last thing we have to do to get delta T prime is take the screw root. And that's gonna be 1.766 repeated multiplied by 10 to the exponent negative five seconds. So this is the delta T prime value that we needed to find. Hey, that's the time between events in the S prime frame. Now we can go back to our star equation. OK. So let's go up and remind ourselves here, we have this star equation V is equal to delta X prime divided by delta T prime. We just found delta T prime. We know that delta X prime is 2800 m. So let's substitute in those values. And so OK, so we are going back to equation star. Now we get that V is going to be equal to 2800 m divided by 1.766 repeated multiplied by 10 to the exponent negative five seconds. If we work this out, we get that V is 1.5849 multiplied by 10 to the exponent 8 m per second. OK. So we've now found the speed V of frame S crime with respect to frame S. What we're left to do is write this in terms of the speed of light C. OK. That's how the answer choices are written in terms of C. So we have our speed, we need to write it in terms of C. So what we're gonna do is we're gonna take this value. We have 1.5849 multiplied by 10 to the exponent 8 m per second. And we're gonna multiply that and we know that the speed of light C is equivalent to three multiplied by 10 to the exponent 8 m per second. So we multiply by one C divided by three multiplied by 10 to the exponent 8 m per second. The unit of meters per second will divide out and we're gonna be left with the 0.52 83. OK. And so now we have the velocity of frame S prime with respect to S in terms of C is 0.5283. If we round this to three decimal places, we can see that this corresponds with answer choice B 0.528 C. Thanks everyone for watching. I hope this video helped see you in the next one.

Two events in reference frame S occur 10 μs apart at the same point in space. The distance between the two events is 2400 m in reference frame S'. What is the velocity of S' relative to S?

Key Concepts

Lorentz Transformation

Time Dilation

Relative Velocity

Watch next